Positive rational numbers with sum of their squares equal to $n$ 
Prove that for all positive integers $n$ there exist $n$ distinct, positive rational numbers with sum of their squares equal to $n$.

For $n = 1$ we can just take $1$. For $n = 2$ we can take $\left(\dfrac{1}{5}\right)^2+\left(\dfrac{7}{5}\right)^2 = 2$. How can we generalize this to any $n$?
 A: Note:
$$x^2+y^2=2$$
has infinitely many positive rational solutions. 
Now, if $n=2k$, pick $k$ distinct solutions $(x_1,y_1),..,(x_n,y_n)$, which are not $(1,1)$. Since the solutions are distinct and positive, the set $\{ x_1,.., x_k, y_1,..,y_k\}$ consists of $2k$ distinct numbers.
Now we have
$$x_1^2+..+x_k^2+y_1^2+...+y_k^2=2k=n$$.
If $n=2k+1$, then repeat the above and write:
$$x_1^2+..+x_k^2+y_1^2+...+y_k^2+1=2k+1=n$$.
A: You can get $x^2+y^2=1$ infinitely many ways using Pythagorean triples, just make sure all the triples are distinct, then add the expressions. If $n$ is odd, use an initial $1$ not from a triple.
Edit 2: (previous idea didn't work) Suppose that
$$x=\frac{k^2-2k-1}{k^2+1},\ y=\frac{k^2+2k-1}{k^2+1}. \tag{1}$$
Then $x^2+y^2=2,$ and if $k$ is even both $x,y$ in (1) can be shown to be in lowest terms. This means we have infinitely many distinct solutions to $x^2+y^2=2,$ so that as @N.S suggests in his comment below we can get $n$ distinct rational squares with sum $n.$
A: This question was asked on JEMC 2016 (question 3). 
Here you can find two solutions: the one with induction, as you stated in your question, and the proof that there exist infinitely many $x,y\in\mathbb{Q}^+_0$ with $x^2 + y^2 =2$.
Personally, I find the latter the most comprehensive (it's the solution I came up with when I solved the problem).
